ENGINEERING MECHANICS - MREC Academics

Malla Reddy Engineering College

(Autonomous)

(An Autonomous institution, Autonomy granted by UGC and affiliated to JNTUH, Accredited by NAAC with ‘A’

Grade, Accredited by NBA (2008-11) & Recipient of World Bank Assistance

under TEQIP phase – II S.C.1.1for the period (2011-14))

Maisammaguda, Dhulapally (Post. Via. Kompally), Secunderabad – 500 100.

LAB MANUAL

ENGINEERING MECHANICS

I B. TECH- I SEMESTER

Subject Code: 80305

Academic Year 2018-19

Regulations: MR18

NAME: __________________________________________

USN : __________________________________________

BATCH: _______________SECTION:_________________

2018-19

Onwards

(MR-18)

MALLA REDDY ENGINEERING COLLEGE

(Autonomous)

B.Tech.

I Semester

Code: 80305 ENGINEERING MECHANICS LAB

L T P

Credits: 1.5 - - 3

Prerequisites: Engineering Physics

Course Objectives: Student will be able to learn and understand the various basic concept

and principles of Mechanics like Force & Resultant.

List of Experiments

1. Triangle law and polygon law of forces apparatus.

2. Support reaction for a beam apparatus.

3. Coefficient of friction apparatus.

4. Flywheel apparatus.

5. Moment of disc apparatus.

6.Jib Crane

7. Equilibrium of non concurrent forces apparatus.

8. Universal force table.

9. Screw jack apparatus.

10. Compound pendulum.

11. Worm and worm wheel apparatus.

12. Differential wheel and axle apparatus.

Course outcome:

At the end of the course, students should be able to:

Course Outcome

Level of

learning

Bloom’s

Taxonomy

Level

CO1 Verify law of Force Polygon and law of Moments using Force

Polygon and bell crank lever apparatus and also study Parallel

Force apparatus.(Simply supported type).

3 Analyzing

CO2 Determine mechanical advantage, Velocity ratio and efficiency of 2 Analyzing

a screw jack.

CO3 Evaluate co-efficient of friction between trolley (slider) and an

inclined plane.

3 Analyzing

CO4 Verify the error percentage for the universal force table. 3 Analyzing

CO5 Determine the moment of inertia of the fly wheel. 2 Analyzing

INDEX PAGE

Sl. Dates

Name of the Experiments

No

Conduction Repetition

Submission of Record

Man

ual M

ark

s

Re

co

rd

Ma

rks

Sign

atu

re(S

tude

nt

) Sig

nat

ure

(Fac

ult

y)

Average

Note: If the student fails to attend the regular lab, the

experiment has to be completed in the same week. Then the

manual/observation and record will be evaluated for 50% of

maximum marks.

II

General Instructions to Students

Students are informed to present 5 min before the commencement of lab.

Students must enter their name in daily book before entering into lab.

Students must leave Foot wares before entering lab.

Students must not carry any valuable things inside the lab.

Students must inform lab assistant before He/She uses any computer.

Do not touch anything with which you are not completely familiar. Carelessness may

not only break the valuable equipment in the lab but may also cause serious injury to

you and others in the lab.

For any software/hardware/ Electrical failure of computer during working, report it

immediately to your supervisor. Never try to fix the problem yourself because you could

further damage the equipment and harm yourself and others in the lab.

Students must submit Record book for evaluation before the commencement of lab.

Students must keep observation book (if necessary).

Students must keep silent near lab premises.

Students are informed to follow safety rules.

Students must obey lab rules and regulations.

Students must maintain discipline in lab.

Do not crowd around the computers and run inside the laboratory.

Please follow instructions precisely as instructed by your supervisor. Do not start the

experiment unless your setup is verified & approved by your supervisor.

CONTENTS

Sl. No.

Title

1. Triangle law and polygon law of forces apparatus.

2. Support reaction for a beam apparatus.

3. Coefficient of friction apparatus.

4. Flywheel apparatus.

5. Moment of disc apparatus.

6.Jib Crane

7. Equilibrium of non concurrent forces apparatus.

8. Universal force table.

9. Screw jack apparatus.

10. Compound pendulum.

11. Worm and worm wheel apparatus.

12. Differential wheel and axle apparatus.

Screw Jack Apparatus of 67

ENGINEERING MECHANICS LAB I SEM


Instruction Manual

Aim :-

To verify the polygon law of forces.

Apparatus :-

1. Gravesand’s Apparatus,

2. Paper Sheet,

3. Weight Box,

4. Thread,

5. Drawing pins & Pencil,

6. Mirror Strip Pans.

Theory :-

“Polygon law of apparatus” states that if a number of forces acting on a particle are

represented in magnitude & direction by sides of a polygon taken in same orde, then their

resultant is represented in magnitude and direction by the closing side of the polygon taken in

the opposite direction.


Procedure :-

1. Set the board in a vertical plane & fix the paper sheet with drawing pins

2. Pass a thread over two pulleys

3. Take a second thread & tie the middle of this thread to the middle of first thread

4. Pass the ends of second thread over the other set of two pulleys

5. Take a third thread & tie its one end to the point of first two threads

6. Attach pans to the free ends of the threads

7. Place the weights in the pans in such a manner that the knot comes approximately in

the centre of the paper.

8. Mark the line of forces & write down the magnitude of forces

9. Remove the paper from the board & produce the line to meet at centre point O

10. Select a suitable scale & draw the vector diagram by moving in one direction. draw a b

parallel to A B & cut it equal to force P; draw b c parallel to B C & cut it equal to Q;

draw c d parallel to C D & cut it equal to force R; draw d e parallel to D E & cut it equal

to force S. vector a e will be the resultant force T1 taken in the opposite direction &

should be equal to force T which proves the law of polygon forces. If a e is not equal to

T then percentage error is found as follows.

Observations :-

S. NO.

FORCES

( total weights of pans)

CALCULATED

RESULTANT T1

PERCENTAGE ERROR

T – T1 x 1000 100


Precautions :-

1. Pans/weights should not touch the board

2. There should be only one central knot on the thread which should be small

3. While calculating the total force in each case the weight of the pan should be added to

the weight onto the pan

4. Make sure that all pans are at rest when the lines of action of force sare marked

5. All the pulleys should be free from friction

I N S T R U C T I O N M A N U A L

P Q R S T

1.

2.

3.


Parts Details

1 Wooden Base 6 Sliding Hook Brackets

2 Iron Strip 7

Extinction Balance With

Chain

3 Upright Bar 8 Hanging Chain With Hook

4 Cast Aluminum Bracket 9 Nut & Washer

5

Tie Bar With Compression

Balance 10

Screw For Fixing The Iron

Strip


AIM : To calculate the stresses in various member of the jib crane

and find the % age error between the calculated and the observed

values.

Things required :: Jib Crane Apparatus

Weights

Procedure ::

1. Note the zero error in the compression balance and the spring

balance.

2. Suspend a wt (w) from the pt. A. Then measure AB,

BC, AC, Draw the scale. Note. B is center of the

compression balance and C s vertically above B. Fig.

can serve as the vector or stress diagram also because it is a

triangle (as any other triangle with sides parallel to it will be a

similar triangle.

3. Let w be represented by CB. Then BA will be represent stress in

the Arm AC (Tie). Thus these are the calculated values (Note, we can

do without the scale diagram and simple measure AB, AC and BC)

4. Note the reading in the compression balance which will give the

observed stress in the Arm AB, and tension in the spring balance

given the observed force in the Arm AC. Find the %age error

between the calculated and observed value.


5. We have to know whether arms AB and AC are in compression or

in tension. Now the pt. A is in equilibrium acted upon by the forces

exerted by arms AB AC and wt (w) as show in FIG 3. Lets first

consider the nature of the force in the arm AB. The opposite force is

exerted by the pt A on the member AB. As the member AB is in

equilibrium another opposite force is exerted by the pt Bon the

member AB. Thus AB is in compression. Similarly it can be show

that the arm AC is in tension i.e. it is stretched.



AIM:-

To determine the efficiency of a worm and worm wheel apparatus

and plot a graph between W & P and W & ŋ.


WORKING:-

As the pulley of the worm moves through n revolutions, n teeth of the wheel pass completely through the worm. If there are N teeth in the worm wheel, then N revolutions will have to be given to the

pulley of the worm to rotate completely the worm wheel. So that if the effort applied at the pulley of the worm moves through N

revolution the load is raised up by the a distance equal to the length of the circumference of the pulley of the worm wheel.

So that velocity ratio is

N x circumference of pulley of worm Circumference of pulley of worm wheel

SINGLE THREADED WORM :- Specifications : Starts : Single

Lead angle : 3° 35'

Pressure angle : 24°


Normal Axial Pitch : 8mm

Pitch of two continuous threads : 21.5mm THEORY :- If the worm is single threaded (i.e for one revolution of the wheel, the worm pushes the worm wheel through one teeth) then for one revolution of the effort pulley the distance moved by effort load = 2пl

The load drum will move through 1 revolution T

Therefore, distance moved by the load on load drum = 2пr T And velocity ratio(V.R) = distance moved by P

distance moved by W V.R = 2пl = lT 2пr r T Mechanical advantage = W P Efficiency, ŋ = M.A V.R


PROCEDURE:-

1. Wrap the string round the pulley of the worm the free end of

which is to be tied to the effort.

2. Wrap another string to carry the load round the pulley the worm

wheel in such a manner that as the effort is applied the load is lifted

up.

3. Suspend a small weight (w) through the free end of the second

string and suspend another weight (p) through the free end of the

first string which should just move the load upward.

4. Note w and P, so that mechanical advantage is given by W

P

5. Increase the load (w) gradually and increase the effort (p)

correspondingly and take in this way about seven readings.

6. Measure the circumference of the pulley of the worm and also

that of the worm wheel.

7. The percentage efficiency is given by W x 100

PV

8. Plot a graph between w and p and w and.

OBSERVATION:-


1. Circumference of pulley of the worm (2∏R1) = 39.25 cm.

2. Circumference of pulley of the worm wheel (2∏R2) = 43.96 cm.

3. No of teeth in the worm wheel (N) = 120

4. Velocity ratio = N x 2πR1 = 107.14

2πR2

OBSERVATION TABLE

S. No. Weight Lifted (W)

Effort Applied (P) Mech. Adv. = W/P

Velocity Ratio

% Efficiency W/PV x 100

1 2 180 107.14

2 3 200 107.14

3 4 250 107.14

4 5 300 107.14

5 6 350 107.14

PRECAUTIONS:-


1. Lubricate the apparatus.

2. There should be no overlapping of the strings.

3. W should move slowly and in upward direction.

4. Effort should be applied carefully on sides of hangers.

INSTRUCTION MANUAL

Aim:

To determine:

The acceleration g of gravity using a compound pendulum.

The radius of gyration kG of the compound pendulum about an axis perpendicular to the

plane of oscillation and passing through its centre of mass.

The moment of inertia IG of the compound pendulum about an axis perpendicular to the

plane of oscillation and passing through its centre of mass.

Apparatus: Bar Strip Length: 1 mtr

Bracket Weight (Black colour): 600 gm

Bar Pendulum Weight: 1.5 kg


Width: 5 mm

Thickness: 2.5 cm

Theory:

In Fig. 1, O is the point of suspension of the compound pendulum

and G is its centre of mass; we consider the force of gravity to be

acting at G. If h is the distance from O to G, the equation of motion

of the compound pendulum is

Where I0is the moment of inertia of the compound pendulum about

the point O.

Comparing to the equation of motion for a simple pendulum

We see that the two equations of motion are the same if we take

It is convenient to define the radius of gyration k0 of the compound pendulum such that if all the

mass Mwere at a distance k0 from O,

the moment of inertia about O would be I0, which we do by writing I0 = Mk02

Substituting this into (1) gives us

The point O′, a distance l from O along a line through G, is called the center of oscillation. Let h′

be the distance from G to O′, so thatl=h+h'. Substituting this into (2), we have

If IG is the moment of inertia of the compound pendulum about its centre of mass, we can also

define the radius of gyration kG about the centre of mass by writing IG = MkG2.

The parallel axis theorem gives us

Comparing to (3), we have,

If we switch h with h′, equation (4) doesn’t change, so we could have derived it by suspending

the pendulum from O′. In that case, the center of oscillation would be at O and the equivalent

simple pendulum would have the same length l. Therefore the period would be the same as when

suspended from O. Thus if we know the location of G, by measuring the period T with

suspension at O and at various points along the extended line from O to G, we can find O′ and

thus h′.

Then using equation (4), we can calculate kG and IG = MkG2.


Knowing h′ gives us l = h + h′, and since for

small angle oscillations the period

We can calculate g using

The minimum period Tmin , corresponds to the

minimum value of l. Recall that l = h + h′ and that kG2 = hh' is a constant, depending only on the

physical characteristics of the pendulum.

Thus, l=h+kG2/h, and the minimum I occurs when,

i.e, when h2=kG2, h=h' and l=2h=2kG.

Thus, at Tmin, l=2kG.

Performing the real lab:

The compound bar pendulum AB is suspended by passing a knife edge through the first

hole at the end A. The pendulum is pulled aside through a small angle and released,

whereupon it oscillates in a vertical plane with a small amplitude. The time for 10

oscillations is measured. From this the period T of oscillation of the pendulum is

determined.

In a similar manner, periods of oscillation are determined by

suspending the pendulum through the remaining holes on

the same side of the centre of mass G of the bar. The bar is

then inverted and periods of oscillation are determined by

suspending the pendulum through all the holes on the

opposite side of G. The distances d of the top edges of

different holes from the end A of the bar are measured for

each hole.The position of the centre of mass of the bar is

found by balancing the bar horizontally on a knife edge.

The mass M of the pendulum is determined by weighing

the bar with an accurate scale or balance.

A graph is drawn with the distance d of the various holes

from the end A along the X-axis and the period T of the

pendulum at these holes along the Y-axis. The graph has

two branches, which are symmetrical about G. To determine

the length of the equivalent simple pendulum corresponding


to any period, a straight line is drawn parallel to the X- axis from a given period T on

the Y- axis, cutting the graph at four points A, B, C, D. The distances AC and BD,

determined from the graph, are equal to the corresponding length l. The average length l =

(AC+BD)/2 and l/T2 are calculated. In a similar way , l/T2 is calculated for different

periods by drawing lines parallel to the X-axis from the corresponding values of T along

the Y- axis. l/T2 should be constant over all periods T, so the average over all suspension

points is taken. Finally, the acceleration due to gravity is calculated from the equation g=

4π2(l/T2).

Tmin is where the tangent EF to the two branches of the graph crosses the Y-axis. At Tmin,

the distance EF = l = 2kG can be determined, which gives us kG, the radius of gyration of

the pendulum about its centre of mass, and one more value of g, from g= 4π2(2kG/Tmin2) .

kG can also be determined as follows. A line is drawn parallel to the Y -axis from the

point G corresponding to the centre of mass on the X-axis, crossing the line ABCD at P.

The distances AP = PD = AD/2 = h and BP = PC = BC/2 = h′ are obtained from the

graph. The radius of gyration kG about the centre of mass of the bar is then determined by

equation (4). The average value of kG over the different measured periods T is taken, and

the moment of inertia of the bar about a perpendicular axis through its centre of mass is

calculated using the equation IG=MkG2.

Performing The Simulation:

Suspend the pendulum in the first hole by choosing the length 5 cm on the length slider.

Click on the lower end of the pendulum, drag it to one side through a small angle and

release it. The pendulum will begin to oscillate from side to side.

Repeat the process by suspending the pendulum from the remaining holes by choosing

the corresponding lengths on the length slider.

Draw a graph by plotting distance d along the X-axis and time period T along the Y-axis.

(A spreadsheet like Excel can be very helpful here.)

Calculate the average value of l/T2 for the various choices of T, and then calculate g as in

step 2 above.

Determine kG and IG as outlined in steps 3 and 4 above.

Repeat the experiment in different gravitational environments by selecting an

environment from the drop-down environment menu. If the pendulum has been oscillating,

press the Stop button to activate the environment menu.

Observations:


To draw graph:

1. To find the value of 'g' :

2. To find the radius of gyration and the acceleration of gravity (step 3 above):

Radius of gyration about the centre of mass kG = EF/2 = ..................

Acceleration of gravity, g= 4π2(2kG/Tmin2) = ......................

3. To find the radius of gyration (step 4 above):


Results:

Average acceleration of gravity, g= 4π2(l/T2) = .................. m/s2

Average radius of gyration of the pendulum about its centre of mass, kG = .................. m

Mass of the pendulum M = .................. Kg

Moment of inertia of the pendulum about its centre of mass, IG=MkG2 = ..................

Kgm2


IN

S T R U C T I O N M A N U A L

AIM:-

Determination of Coefficient of friction by the inclined plane apparatus.


APPARATUS:- 1. Inclined plane

2. Sliding boxes with different surfaces 3. String

4. Pan

5. Thread

THEORY:-

When a body slides upon another body, the property by virtue of which

the motion of one relative to the other is related is called friction. The frictional force is directly proportional to the normal reaction ‘N’.

F ∞ N


F = μN or μ = F

N

Suppose a body of weight W is to be lifted by inclined plane & this requires effort P. when this load just move upward a frictional force F

acts downward which oppose its motion.

Component of load W parallel to the plane = W sin

Component of load W perpendicular to the plane = W cos

Considering equilibrium parallel to the plane

P = F + W sin

F = P – W sin ……(i)

Considering equilibrium perpendicular to the plane

N= W cos ……….(ii)

From (i) & (ii)

Co – efficient of friction,


μ = F = P – W sin

N W cos

Mechanical advantage (M.A.) = W P

Velocity ratio (V.R.) = distance moved by effort Distance moved by load

Let effort P comes down through one centimeter, movement of the load along the plane = 1 cm

Vertical uplift of load = 1 x sin

V.R. = 1 = cosec

1 x sin

% efficiency = M.A. x 100

V.R.


PROCEDURE:-

1. Take the inclined plane apparatus & keep it first horizontal and

put the slider on it.

2. Increase the inclination of inclined board gradually till the slider

just begins to slide downwards on it.

3. Note the angle in this position. This is called angle of repose.

4. Place the slider on the plane with the desired angle .

5. Tie the slider to the pan with the help of thread passing over the

pulley.

6. Put the weight in the pan till the slider just start moving. note

down the weight.

7. Measure the angle of inclination from the scale provided & finds the value of μ.

8. Calculate M.A., V.R., efficiency.


OBSERVATIONS:-

S. No.

Total weight of

the slider W (grams)

Weight of pan +

weight in the pan

P (grams)

μ = P – W sin W cos

M.A. = W P

V.R.= cosec

% Efficiency

1.

125

40

0.15

3.125

5.75

54.34

2.

125

65

0.35

1.92

5.75

33.39

3.

125

115

7.65

1.08

5.75

18.72

4.

150

80

0.37

1.87

5.75

32.6

5.

200

110

0.38

1.81

5.75

31.62

6.

150

67

0.28

2.23

5.75

38.93

7.

200

115

0.41

1.73

5.75

30.24

8.

350

155

0.27

2.25

5.75

39.27

9.

150

62

0.24

1.73

5.75

30.24


PRECAUTIONS:-

1. The plane should be clean & smooth.

2. The guide pulley should move freely. It should be lubricated to

make it frictionless.

3. Weight should be added gently in pan.

4. The slider should just begin to move slowly, it should not move

abruptly.

5. The direction of thread should be parallel to the inclined plane.


MOMENT OF INERTIA OF FLYWHEEL

APPARATUS

CONTENTS: Page No.

AIM 02

APPRATUS USED 02

DESCRIPTION 03

PROCEDURE 04

MATHEMATICAL CALCULATION 05

OBSERVATION 07

PRECAUTIONS 08



AIM: - To find the moment of Inertia of a Fly Wheel.

APPARATUS USED:-


1. Fly wheel

2. Meter scale

3. Vernier calipers

4. Stop watch

5. Weight

DESCRIPTION OF APPARATUS:- A flywheel is a large heavy wheel, through the center of which

passes a long cylindrical axle. The center of gravity lies on its axis of rotation so that when it is mounted over ball bearing, it comes to

rest any desired position.

To increase the moment of inertia, it is usually made thick at the

rim.

To count the number of revolution made by the wheel, a lime is

marked on the circumference. A string is wound on the axle, attached to the peg carries a mass M.

THEORY :

When a weight is suspended to the free end of a cotton string which

is wrapped round tin: shaft and the other end of which is tied to the

shaft and it is allowed to fall to touch the ground then P.E. energy

possessed by the falling weight has partly been used to give motion

to the fly wheel and partly used in overcoming frictional resistance

present in the bearings.


1. Initial Potential Energy (P.E.)=Wxh when h is the height of the

weight from the level ground.

2. Initial Kinetic energy (K.E.) = 0. Final P.E = 0.

3. Final (K.E) = ½ (w/g) v2 + ½ I (w2 / g) where ½ (w/g) v2 is the K. E

of the falling wt. And ½ I (w2/g) is the K. E. of the fly wheel and

shaft combined and w is the final angular velocity of the wheel or of

the shaft.

4. Work done due to frictional resistance = F x h where F is

the force of friction acting tangentially to the shaft.

From law of conservation of Energy we can write

W x h = ½ (w/g) v2 + ½ I (w2 / g) + F x h

=> 2g(W - F) x h = W v2 + I v2

=> I = (W – F) x 2gh - W v2

w2

But v (final velocity) is given by the formula (u+v)/2 = h/2 where t is the time taken for the weight to fall a distance h so that h/t is the

average velocity.

PROCEDURE:-

1. Attach a mass M (about 500 grams) to one ned of the thin thread and loop is made at the other end which is fastened at

the peg.

2. The thread is wrapped evenly round the axle of wheel.

3. Allow the mass to descend slowly & count the number revolution N1 during descent.


4. When the thread has unwound itself & detached from the axle after N1 turns, start the stop watch. Count the number

revolution before the flywheel comes to rest & stop the stop watch. Thus N2 are known.

5. With the help of vernier calliper, measure the diameter at several

point. Thus find R.

6. Repeat the experiment with three

different masses.

7. Calculate the value of I using the given formula.

MATHEMATICAL CALCULATIONS :

Moment about axis ox,

Consider an elementary ring of radius ‘x’ and thickness dx

Moment of inertia of this elementary ring about axis OX 2

= mass x (radius) 2

3

= pпx dx Taking mnt of inertia about OX ie integrating the above

equation 2

Iox = Mr


4 2

By symmetry Ioy = M (r) 4

Similarly for axis OZ 2

M.O.I = mass x (radius) Taking mnt about OZ 4

Ioz = pпr

2 2

= Mr 2

OBSERVATIONS:-

S.n

o

Total load

applied M (k.g.)

No. of

revolutions of flywheel before

the mass deattached

N1

No. of

revolutions of flywheel

to come to rest after

mass

deatached N2

Mean

N2

Times of N2

revolution T seconds

Mean T

sec.

1.

o.5

12

53 59 62

58

1.25 1.37 1.38

1.33

2.

1

12

107 120 121

116

2.05 2.12 2.23

2.13


3.

2

12

157 161

158

158

2.17 2.23

2.20

2.20

PRECAUTIONS:- 1. Note the time accurately to the fraction of a second

2. Note the value of F when the motion just begins and the fly wheel does not move with any acceleration

3. Oil the bearings to reduce friction

4. Overlapping of the string should be avoided

5. Note the time thrice for the same weight (W)


MOMENTS DISC APPARATUS

CONTENTS: Page No.

OBJECTIVE 03 THEORY 04 PROCEDURE 05 OBSERVATION 06 PRECAUTIONS 07



AIM: -

To verify the law of moment by rotating disc apparatus.

APPARATUS REQUIRED:-

1. Mirror scale (adjustable) 2. Two small pulley (adjustable 3. Hollow disc (adjustable)


4. Four pans 5. Slotted weight 6. Plumb line

THEORY & FORMULA USED:-

The law of moment’s states that if a number of coplanar forces acting on a rigid body keep it in equilibrium then the algebric sum of their moments about any point in their plan is zero.

In the moments disc apparatus, we use the hollow disc, pulleys, threaded pan, mirror scale, plumb line. All these things are adjustable. Due to use these apparatus make to ensure that the level of fixing threaded pan in a hollow disc should be equal. Requirement of plumb line to adjusting the apparatus at equal level in any positions. The hollow disc & pulleys are moving either clockwise or anti-clockwise direction. All parts are adjusting on the straight beam which should be rigidly fixed on the base of the apparatus.


PROCEDURE:-

(1) Put weights in the two pans A and B. such that W1 is the weights in the pan A plus weight of the pan & W2 weight in the pan B plus weight of pan B. Note down W1 & W2 .

(2) Rotating disc should be placed at a centre point note the thread are showed at zero on the scale

(3) Note down the distance X1 and X2.

(4) Now ( W1 x X1 ) will be equal to (W2 x X2 ) that is both the clockwise and anticlockwise and anticlockwise moments will be equal.

(5) Take different sets of reading and find out the value of both the moments.


(6) If both the moment is not equal then find out the percentage error between the clockwise moment and the anti clockwise moment.

OBSERVATIONS:- W1 = weight in the pan A W2 = weight in the pan B X1 = distance from the centre point of the apparatus to the end of thread

shadow show in the mirror scale of which pan A X2 = distance from the centre point of the apparatus to the end of thread

shadow show in the mirror scale of which pan B


S. No. Weight of pan + Weight in pan (W1)

Weight of pan + Weight in pan (W2)

X1 (c.m.)

X2 (c.m.)

Percentage Error

1.

225

225

17

17

0

2.

225

225

17

16.6

0.4

3.

525

525

17

16.6

0.4

PRECAUTIONS:- (1) Weights should be placed in the pans A and B gently. (2) Take in to the account the weights of the pan. (3) Lubricate the apparatus. (4) Distance should be noted down carefully.


Parallel forces Apparatus


CONTENTS: Page No.

AIM 03

THINGS REQUIRED 03

APPRATUS 03

THEORY 03

NOMENCLATURE 04

PROCEDURE 05

OBSERVATION 05

PRECAUTIONS 06



AIM:

To verify the principle of forces in beam of Parallel Forces Apparatus with the help of beam

supported at its ends.

Things required:

Parallel Forces Apparatus 10 Kg Dial Type, Conical Weights, and Aluminum hangers for

hanging weights.

Appratus :

The apparatus comprising of two dial type weight gauges of 10 kg, one straight wooden beams

of 150cm, a wooden platform for the support of the dial gauges, Two weight hangers for

hanging the weights on the wooden beams, Two weights weighing 2 kg aggregate. The beam is

provided with angular slots on them in order to place the hanger in it, the distance between

each groove is 5cms. The weight of each hanger will be neglected.

The whole apparatus is well designed & painted.

Theory:

If a system of coplanar forces acting on a rigid body keep it in equilibrium then the algebraic

sum of their moments about any point in their plane is zero. Normally a beam is analysed to

obtain the maximum stress and this is compared to the material strength to determine the

design safety margin. It is also normally required to calculate the deflection on the beam under

the maximum expected load. The determination of the maximum stress results from producing

the shear and bending moment diagrams.


The sign convention used for shear force diagrams and bending moments is only important in

that it should be used consistently throughout a project. The sign convention used on this

page is as below.

Nomenclature :

e = strain

σ = stress (N/m2)

E = Young's Modulus = σ /e (N/m2)

y = distance of surface from neutral

surface (m).

R = Radius of neutral axis (m).

I = Moment of Inertia (m4 - more normally cm4)

Z = section modulus = I/ymax(m3 - more normally cm3)

M = Moment (Nm)

w = Distrubuted load on beam (kg/m) or (N/m as force units)

W = total load on beam (kg ) or (N as force units)

F= Concentrated force on beam (N)

S= Shear Force on Section (N)

L = length of beam (m)

x = distance along beam (m)

Procedure:

1. First of all arrange the apparatus by placing the beams on the given dial gauges as

shown in the figure. Note the zero error in the compression balances. When the beams

are supported at it ends.

2. Suspend three different weights

from the sliding hook against any

division marked on the beam.

3. Note the reaction on the beam

given by the readings of

compression balances and takes

into account the zero error from

each reading.

4. Find the some of clockwise moment about the mid points of the beam and find also the

sum of anti-clockwise moment about its each reading.

5. Find the % age error between clockwise and anti-clockwise moment.


6. Suspend three or four weights at different graduated division of the beam and find %age

error between clockwise and anti –clockwise moments as before.

Observations:

1. Zero error in compression balance No. 1= -------------

2. Zero error in compression balance No. 2= -------------

TABLE:

S.no Weight W1 Weight

W2

Reaction

R1

Reaction

R2

Reaction R3 Distance x1 Distance

X2

Take in this manner about seven readings.

Precautions :

1. Zero error of the compression balances must be taken in to account,

2. Weights should not be put on the beam with a jerk.

3. Slightly press h beam in to remove any frictional resistance at the supports before

taking readings.


CONTENTS: Page No.

Objective 02 Accessories Required 02 Principle 02 Theory 02 Procedure 03 Formula’s 03 Nomenclature 04 Observation Table 04 Calculations 04 Results 05 Precautions 05

Instruction Manual

SCREW JACK APPARATUS AIM:-

To determine the mechanical advantage, velocity ratio and efficiency of a screw jack.


ACCESSORIES REQUIRES:-

Screw Jack

Thread/Rope

Load Weight (5kgs)

Weight Set (50gms -2, 100gms-2)

1 measuring Ruler/Inch Tape


PRINCIPLE:

Basically screw jacks are mechanical devices consisting of a heavy bottom metallic base or stand

through which a screw mechanism is allowed to slide up and down through a circular path over a central

axis. The load that is to be lifted is placed over the top “head" of the screw mechanism. The lifting

movement or operation is made functional by applying an external physical force (using human hands)

through a radial motion. A careful inspection of the screw movement (unwinding) through a single thread

shows that the elevating movement follows the principle of an inclined plane.

Primarily two major factors are involved with the functioning of a screw jack, viz.

the weight lifted and the effort applied.

THEORY:-

Screw Jacks are devices used for raising and lowering heavy objects (weights) through external

small manual labor. The article discusses the various calculations involved and explains the relation

between weight which is to be lifted and the applied external force associated with these devices. Screw

jacks were especially developed with the aim of solving the issue of raising and lowering large and heavy

objects over ground levels easily through relatively smaller magnitudes of external effort. Screw jack

works on the principle of screw & nut. It is used for raising heavy loads through small efforts.

A simple screw jack consists of a nut, a screw square threaded and a handle fitted to the head of the screw.

The nut also forms the body of the jack. The load to be lifted is placed on the head of the screw. Here the

axial distance between corresponding points on two consecutive threads is known as pitch. If ‘ps’ be the

pitch of the screw and‘t’ is the thickness of thread, then ps = 2t.

PROCEDURE:-

Step 1 Wrap on string round of the flanged table and take it over a small pulley P1. Effort load P1 is tied to

the free end of this string.

Step 2 Wrap another string around of the flanged table in the same direction in which the above string is

wounded and take it over the other small pulley P2.

Step 3 Place a load ‘W’ on the flanged table.

Step 4 Suspend weights over P1 and P2 to the free ends of both the strings coming over the two small

pulleys. This load P1 and P2 should be increased gradually suitable to give a just upward motion

to the load ‘W’ placed on the screw head.

Step 5 Note down ‘W’ and ‘P’ to determine the M.A. i.e W/P. in this case effort weight would be added and

it consider as total effort P; P1+P2.

Step 6 Note down the distance fall down to lift the load l, measure by ruler.

Step 7 Note down the distance covered by flanged table from initial to final, measure by ruler.

Step 8 P1 and P2 should be noted down carefully.

Step 9 Take 3-4 reading by the variation of the Loads and Efforts.

FORMULA’S:

1. Circumference of the screw thread, = π*D

= 3.14 * 3.05 (π = 3.14, D = 3.05 cm)

= 9.5

2. Mechanical Advantage = Load/Total Effort


3. Velocity Ratio = Distance moved by the effort/Distance moved by the load

4. Efficiency:

NOMENCLATURE:

D : Diameter of Screw

ps : Pitch of the screw

W : Load suspended on flanged table

P : Total Effort Applied

P1 : Weight suspended pulley 1

P2 : Weight suspended pulley 2

l : Distance fall down to lift the load, i.e to turn the screw.

R : Distance covered by flanged table in upward direction while weight

suspended via pulleys (By Ruler)

OBSERVATIONS:-

Diameter of the rotational centre screw, D = 30.5 mm = 3.05 cm Pitch of the center long screw, ps = 2.5 mm = 0.25 cm Diameter of flanged table, = 205 mm = 20.5 cm Thickness of the flanged table = 41 mm = 4.1 cm No. of Pulley’s, P1, P2 = 2 Diameter of pulley = 52.5 mm = 5.25 cm Thickness of pulley = 10 mm = 1 cm Circumference of the flanged table = 9.5 Load Lifted, W = Weight lifts Total Effort, (P1 + P2), P = Efforts applied

Note: Take P1 & P2 as per applicant need.

CALCULATIONS:-

M.A. = W/P V.R. = Distance moved by effort/Distance moved by load Efficiency,% = (M.A./V.R) X 100

RESULTS: - ……………………………………………………

PRECAUTIONS:-

5. There will be no overlapping of the strings.

6. “W” should moves upward direction gently.

Observations Calculations

Sr. No.

Load Lifted,

W, Gms

Effort P1

Effort P2 Total Effort

P1+P2 R

mm l

mm M.A. V.R

Efficiency

1.

2.


7. Load always place on flanged table.

8. Effort should be applied carefully on both sides of hangers.

9. Both the pulleys are in parallel direction meanwhile performing.

10. Lubricate the screw thread in every 3-4 weeks

11. Grease should be used as lubricant always.

SAMPLE READINGS:

(All measurements constructed in shortest units)

RESULTS: The relationship between the efficiency and the load, the effort against the load as shows that increase in load causes decrease in efficiency and increase in load causes increase in effort applied.

Observations Calculations

Sr. No.

Load Lifted,

W, Gms Effort P1 Effort P2

Total Effort P1+P2

R mm

l mm

M.A. V.R Efficienc

y

1. 5000 400 300 700 10 120 5.55 75.3 9.4

2. 5500 550 850 1400 10 130 3.62 81.68 4.37


UNIVERSAL FORCE TABLE

INSTRUCTION MANUAL

UNIVERSAL FORCE TABLE


AIM:- To Verify The Universal Force Table.

COMPONENTS:-

Thread Pulley Slotted Weight Thread/Rope Level Screw Central Ring


Observation Table:

Calculation Table:

Sr. No

Angles(Ɵ) And Distances Ring To Wheel (cm)

RESULTANT (R)

ɵ1 L1 ɵ2 L2 ɵ3 L3 ɵ4 L4 ɵ5 L5 PYTHA. THEROM

1 0 22.3 72 21 120 20.8 170 22.0 28.8 23.5 49.06

Sr. No

Forces Or Weights Suspended (R1)

F=W1 F=W2 F=W3 F=W4 F=W5 F=W6

1 0 200 200 350 250 250

Sr. No.

(R1) AND (α) Percentage Error In Resultant And Angles TOTAL

%ERROR IN

F= SUM F α

(B-A)

1 1.250

KG 72 97.45 75 22.45


INSTRUCTION MANUAL

Experiment: To determine the efficiency of a wheel and differential axle and to plot

a graph between W and P and (w and n)

Things required:

1. Wheel & Differential Axle Apparatus

2. Weights

3. Cotton Thread

Theory:

When the wheel moves through n revolutions the bigger and smaller axles also move

through “n” revolutions. But while the string carrying weight is wrapped round the

bigger axle completely it is released completely it is released completely from the

smaller axle the same time if the radius of the bigger axle is r1 and smaller axle is r2

then the string of length nr n (2 r1) is raised while a length nr n x 2r2 is lowered so

that weight (w) which is supported by two segments of string is raised through a

distance (2r1- 2r2)x n: while distance moved by the effort is 2R x n

Velocity Ratio (v) is 2r x n = 2(2r) in term of the circumferences

(2r1 - 2r2) x n (2r1-2r2)

Procedure::

1. Wrap the string round the smaller and bigger axles in such a manner that as it

is released from the smaller axle in such a manner that is it released from the


smaller axle it is wrapped round the bigger axle Tie another string round the

wheel which will carry the error applied.

2. Suspend a smaller weight (w) through the hook of a small pulley going in

between the two segment of the string as show in the diagram.

3. Suspend another weight (p) is to act as effort to just raise the load (w)

4. Note the wt (w) and effort P so that mechanical advantage is w/p.

5. Increase the weight (w) by a small amount and go on increasing the effort P

gradually till he load (w) is just raised. Take in this way about seven readings.

6. Measure the circumference of smaller and bigger axles and of the wheel also, so

that velocity ratio (v) is determined.

7. The % age efficiency is given by W X 100

PV

Observations:

1. Circumference of smaller axle (2Πr1) = ----------- cm

2. Circumference of bigger axle (2Πr2) =------------ cm

3. Circumference of the wheel= (2ΠR) =------------ cm

4. Velocity ratio = 2ΠR =------------

(2Πr1- 2Πr2)

Table:

S. No. Weight

Lifted (W)

Effort Applied

(P)

Mech. Adv. =

W/P

Velocity

Ratio

% Efficiency

W/PV x 100

1

2

3

4

5

Precautions:

1. Note the value of F when the motion just begins and the fly wheel does not move with any acceleration

2. Oil the bearings to reduce friction. 3. Overlapping of the string should be avoided.

ENGINEERING MECHANICS - MREC Academics

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